Math 215B HW 1 Solutions Zhengyan Shi Winter 2020 1 Poincare Duality We first note the following relation between cup and cap products: hφ _ ; σi = hσ ^ φ; i ! hσ ^ ·;[M n]i = h· _ [M n]; σi (1) Now consider the sequence of maps n−k h D∗ k H (M; F ) −! Hom(Hn−k(M; F );F ) −−! Hom(H (M; F );F ) ; (2) where definitions of h and D∗ are explained in Hatcher's book. By the relation between cup and cap products, D∗ ◦ h(σ) = D∗(hσ; ·i) = hσ;[M n] _ ·i = ± hσ ^ ·;[M n]i : (3) Hence showing that the pairing is nonsingular is equivalent to showing that D∗◦h is an isomorphism. But since F is a field, h is an isomorphism by UCT, and D∗ is an isomorphism because it is the hom dual of the isomorphism map that appears in Poincare duality. 2 Compactly Supported Cohomology We derive the result by the following chain of isomorphisms ∗ ∗ ∗ Hc (X; G) ' lim H (X; X − Kr; G) ' lim H (X [ 1;X [ 1 − Kr; G) −! −! (4) ' H∗(X [ 1; 1; G) = H~ ∗(X [ 1; G) where the first and fourth isomorphisms are by definition, the second isomorphism is by excision 1 of 1 and the third isomorphism is by contractibility of X [ 1 − Kr for sufficiently large r. 1 Since X [ 1 − Kr is contractible, the interior of X [ 1 − Kr contains the closure of 1 for any sequence of Kr. Excision hence applies. 1 3 Covering Space Smooth Structure We first note that the covering space of a manifold is still a manifold (i.e. the covering space inherits 2 second-countability+Hausdorff) . To get a smooth structure, we simply take a set of charts (φi;Ui) for X with fUig evenly covered, and lift them to charts fU~i;ag in X~ with a labeling the sheets of the covering space. Smoothness of transition maps follows immediately and π is an immersion because it is a local diffeomorphism with the smooth structure defined above. 4 Submanifold Structure n2 n2 (a) View Matn;n(R) as R . Define det : R ! R. det is a polynomial (in particular continuous). −1 −1 Therefore det (U) is open whenever U is open. In particular, det (R − f0g) = GLn(R) is open. k But open subsets of R are automatically submanifolds of the same dimension. Therefore, GLn(R) is a smooth submanifold of dimension n2. −1 (b) SLn(R) = det (1). For an arbitrary A 2 SLn(R), we can compute the directional derivative det A(1 + )n − det A 1 + n + O(2) − 1 rAA = lim = det A lim = n det A = n : (5) !0 !0 n2 Hence, for all A, the map det∗ : T R ! T R (differential/pushforward of det) is surjective. Hence 1 is a regular value, and by the regular value theorem, SLn(R) is a submanifold of codimension 1 n2 in R . T (c) We work instead with O(n) = fM 2 Matn;n(R)jMM = Ig. Since SO(n) is just one of the two path components of O(n), if we can show O(n) is a submanifold, then SO(n) is a submanifold n2 n2 n(n+1)=2 of the same dimension. Now consider the map f : R ≈ R ! R ≈ Symn;n(R) defined by f(A) = AAT . At some arbitrary element A with f(A) = I, we have: T T T T (A + X)(A + X) − AA (tAX + tXA ) T T rX f(A) = lim = lim = AX + XA (6) !0 !0 SA −1 By taking X = 2 for some symmetric S, rX f(A) spans Symn;n(R) for all A 2 f (I). Therefore, n(n+1) I is a regular value of f and O(n) has codimension equal to dim(Symn;n(R) = 2 . Hence, 2 n(n+1) n(n−1) dim SO(n) = dim O(n) = n − 2 = 2 . 2if you want to know the proof, consult for example Lee's monograph on smooth manifolds 2 5 Real Projective Space n n (a) Let x 2 S label the element [x] 2 RP . Collect the components fij([x]) into a matrix f. Then X X 2 X 2 fij([x]) = xixj = fji([x]) fij([x])fjk([x]) = xixk xj = xixk = fik([x]) tr f = xi = 1 : j j i (7) n 2 This means f(RP ) is a subset of C = fm 2 Symn+1;n+1(R)jm = m; tr m = 1g. Now we show f is bijective. For injective, suppose f([y]) = f([x]), then diagonal components give constraints 2 2 yi = xi 8i, implying yi = ±xi. But suppose that yi = xi; yj = −xj, then xixj 6= yiyj leading to a contradiction. Hence, ± needs to be consistently chosen for all i. However, since representatives x; y should only be taken from a half-sphere, we conclude that there is in fact a unique sign choice and f is injective. To show surjectivity, note that for any m with m2 = m and tr m = 1, m = OT ΛO where O 2 SO(n + 1) and Λ = diag(1; 0;:::; 0). By carrying out the matrix multiplication, we P 2 n see that m = f([x]) where xi = O1i. O 2 SO(n + 1) guarantees that i O1i = 1, and x 2 S . This concludes surjectivity. In fact, this computation gives an explicit map g : C ! Sn such that n π ◦ g is the inverse of f. By construction, f ◦ π; g are polynomial functions of x 2 R and hence smooth. By the general theory of smooth manifolds, when we restrict the domain of f ◦ π to a submanifold Sn we get a smooth map. Similarly, when we restrict the range of g to a submanifold Sn, we get a smooth map (it is important that Sn is embedded. Restriction of the range to an n n immersed submanifold generally kills smoothness). Since π : S ! RP is a covering map (hence a local diffeomorphism) and f is bijective, we conclude that f is in fact a diffeomorphism. P 2 2 (b) For every A 2 C, ij Aij = tr A = tr A = 1. Hence C has bounded Euclidean distance from (n+1)2 the origin of R and must be bounded. C is obviously closed since the three defining properties are preserved under taking limits. By Heine-Borel, C is compact. Since compactness is preserved n under diffeomorphisms, RP is compact. π n (c) Suppose ζ −! B is trivial, then there exists a bundle isomorphism Φ : R × B ! ζ. Define a n set of sections σi(x) = (ei; x) where feig is a complete basis in R . Then fσi(x)g inherits linear independence from feig. Conversely, if there is a set of linearly independent sections σi(x): B ! ζ, n P then define Φ : R × B ! ζ via Φ(v; x) = i viσi(x). Fiberwise, this is a linear isomorphism since fσi(x)g is linearly independent. Hence Φ is a bundle isomorphism. 3.